Question

There is no triangle $ABC$ satisfying ${(a+b)}^2=c^2+ab$ and $\sin A+\sin B+\sin C=1+\frac{\sqrt{3}}{2}$.

• A. True
• B. False

Answer 1

The correct option is B False

${(a+b)}^2=c^2+ab$
$\Rightarrow a^2+b^2+2ab-c^2-ab=0$
$\Rightarrow a^2+b^2-c^2=-ab$
Divide both sides by $2ab$ we get
$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=\frac{-1}{2}$
$\Rightarrow \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{-1}{2}$
$\Rightarrow C={180}^0-{60}^0={120}^0$
$\therefore \sin A+\sin B+\sin C=1+\frac{\sqrt{3}}{2}$. (given)
$\Rightarrow \sin A+\sin B+\sin {120}^0=1+\frac{\sqrt{3}}{2}$. from above
$\Rightarrow \sin A+\sin B+\sin ({180}^0-{60}^0)=1+\frac{\sqrt{3}}{2}$.
$\Rightarrow \sin A+\sin B+\sin {60}^0=1+\frac{\sqrt{3}}{2}$
$\Rightarrow \sin A+\sin B+\frac{\sqrt{3}}{2}=1+\frac{\sqrt{3}}{2}$
$\Rightarrow \sin A+\sin B=1$
$\Rightarrow 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=1$
$\Rightarrow 2\sin (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)=1$ since $A+B+C=\pi$
$\Rightarrow 2\cos \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)=1$
$\Rightarrow 2\cos \left(\frac{{120}^0}{2}\right)\cos \left(\frac{A-B}{2}\right)=1$ since $\angle C={120}^0$
$\Rightarrow 2\cos {60}^0\cos \left(\frac{A-B}{2}\right)=1$
$\Rightarrow 2\times \frac{1}{2}\cos \left(\frac{A-B}{2}\right)=1$
$\Rightarrow \cos \left(\frac{A-B}{2}\right)=1=\cos 0^0$
$\Rightarrow \frac{A-B}{2}=0$
$\Rightarrow A-B=0$ or $A=B$
Since $A+B+C={180}^0$
$\Rightarrow A+B={180}^0-{120}^0={60}^0$
From above we have $A=B$
$2A=2B={60}^0$
$A=B={30}^0$
and hence such triangle is possible.